Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 8

Find the values of $\delta$ that correspond to $\varepsilon = 0.5$ and $\varepsilon = 0.1$ for the $\lim \limits_{x \to 2} \frac{4x + 1}{3x - 4} = 4.5$
to illustrate the definition of the precise limit.








From the Definition,

$\text{if }\quad \displaystyle 0 < | x - 2 | < \delta \qquad \text{ then } \qquad \left| \left( \frac{4x + 1}{3x - 4} \right) - 4.5 \right| < \epsilon$

For $\epsilon = 0.5,$

As shown in the graph, we must examine the region near the point $(2, 4.5)$. Notice that we can rewrite the inequality.

$\displaystyle \left| \left(\frac{4x + 1}{3x - 4}\right)- 4.5 \right| < 0.5$

$\displaystyle 4 < \left| \frac{4x + 1}{3x - 4} \right| < 5$

So we need to determine the values of $x$ for which the curve $y = \left( \frac{4x + 1}{3x - 4} \right)$ lies between the horizontal lines $y = 5$ and $y = 4$ as shown in the graph.
Then we estimate the $x$-coordinate by drawing a vertical line at the point of intersection of the curve and the line up to the $x$-axis to get its distance from where the limit approaches so that we form...

$1.91 < x < 2.13 \qquad$ then $\qquad 4 < \left| \frac{4x + 1}{3x - 4} \right| < 5$

The interval of the $x$ coordinates $(1.91, 2.13)$ is not symmetric about $x = 2$, the distance of $x = 2$ to the left end point is $2 - 1.91 = 0.09$ while at the right is $2.13 - 2 = 0.13$.
Therefore, we can choose $\delta$ to be smaller to these numbers to ensure tha we're able to keep within the range of epsilon, let $\delta = 0.09$. Then we
can rewrite the inequalities as follows.

$|x - 2| < 0.09 \qquad$ then $\qquad \left| \left( \frac{4x + 1}{3x - 4} \right) - 4.5 \right| < 0.5$

$\fbox{Thus, if $x$ is within the distance of $0.09$ from $2$, we are able to keep $f(x)$ within a distance of $0.5$ from $4.5$.}$

For $\epsilon = 0.1$,

If we change the value of epsilon $\epsilon = 1$ to a smaller number $\epsilon = 0.1$, then by using the same method above we find that


$\begin{equation} \begin{aligned} \left| \left( \frac{4x+1}{3x-4} -4.5 \right) \right| < 0.1\\ 4.4 < \left| \frac{4x+1}{3x-4} \right| < 4.6 \end{aligned} \end{equation}$


We can estimate the value of $x$ as

$1.98< x < 2.02 \qquad$ then $\qquad \displaystyle 4.4 < \left| \frac{4x+1}{3x-4} \right| < 4.6$

The value of $\delta$ from the right and left of $2$ is the same, $2-1.98 = 0.02$ and $2.02 - 2 = 0.02$

$\fbox{Thus, if $\delta$ is $0.02$, we are able to keep $f(x)$ within a distance of $0.1$ from $4.5$}$